Formatted question description: https://leetcode.ca/all/646.html
646. Maximum Length of Pair Chain (Medium)
You are given n pairs of numbers. In every pair, the first number is always smaller than the second number.
Now, we define a pair (c, d) can follow another pair (a, b) if and only if b < c. Chain of pairs can be formed in this fashion.
Given a set of pairs, find the length longest chain which can be formed. You needn't use up all the given pairs. You can select pairs in any order.
Example 1:
Input: [[1,2], [2,3], [3,4]] Output: 2 Explanation: The longest chain is [1,2] -> [3,4]
Note:
- The number of given pairs will be in the range [1, 1000].
Related Topics:
Dynamic Programming
Similar Questions:
Solution 1. DP
First sort the array in ascending order.
Let dp[i] be the length of maximum chain formed using a subsequence of A[0..i] where A[i] must be used.
dp[i] = max(1, max(1 + dp[j] | pair j can go after pair i ))
// OJ: https://leetcode.com/problems/maximum-length-of-pair-chain/
// Time: O(N^2)
// Space: O(N)
class Solution {
public:
int findLongestChain(vector<vector<int>>& A) {
sort(begin(A), end(A));
int N = A.size();
vector<int> dp(N, 1);
for (int i = 1; i < N; ++i) {
for (int j = 0; j < i; ++j) {
if (A[j][1] >= A[i][0]) continue;
dp[i] = max(dp[i], 1 + dp[j]);
}
}
return *max_element(begin(dp), end(dp));
}
};
Solution 3. Interval Scheduling Maximization (ISM)
// OJ: https://leetcode.com/problems/maximum-length-of-pair-chain/
// Time: O(NlogN)
// Space: O(1)
class Solution {
public:
int findLongestChain(vector<vector<int>>& A) {
sort(begin(A), end(A), [](auto &a, auto &b) { return a[1] < b[1]; });
int e = INT_MIN, ans = 0;
for (auto &v : A) {
if (e >= v[0]) continue;
e = v[1];
++ans;
}
return ans;
}
};
Java
class Solution {
public int findLongestChain(int[][] pairs) {
Arrays.sort(pairs, new Comparator<int[]>() {
public int compare(int[] pair1, int[] pair2) {
if (pair1[0] != pair2[0])
return pair1[0] - pair2[0];
else
return pair1[1] - pair2[1];
}
});
int length = pairs.length;
int[] dp = new int[length];
Arrays.fill(dp, 1);
for (int i = 1; i < length; i++) {
int[] curPair = pairs[i];
for (int j = 0; j < i; j++) {
int[] prevPair = pairs[j];
if (prevPair[1] < curPair[0])
dp[i] = Math.max(dp[i], dp[j] + 1);
}
}
int max = 1;
for (int num : dp)
max = Math.max(max, num);
return max;
}
}